**Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB ,KVS,STET Exam.**

**Q1. If for ∆ABC and ∆DEF, the correspondence CAB ⟷ EDF gives a congruence, then which of the following is not true?**

(a) AC = DE

(b) AB = EF

(c) ∠A = ∠D

(d) ∠B = ∠F

**Q2. In ∆ABC and ∆PQR, AB = PQ, AC = PR and ∠B = ∠Q = 90°. Then, ∆ABC ≅ ∆PQR by the criterion **

(a) SAS

(b) SSS

(c) ASA

(d) RHS

**Q4. A polyhedron has 12 faces and 16 vertices. The number of its edges are **

(a) 20

(b) 24

(c) 26

(d) 32

**Q6. In ∆ABC, AB = 7 cm, BC = 6 cm and CA = 5 cm. Which of the following is correct?**

(a) ∠A > ∠B

(b) ∠B > ∠C

(c) ∠C = ∠A

(d) ∠C < ∠A

**Q8. Which of the following pairs are supplementary angles?**

(a) 54° and 126°

(b) 195° and 126°

(c) 172° and 8°

(d) (1) and (3)

**Q9. The sides of a triangle are 6.5 cm, 10 cm and x cm, where x is a positive number. What is the smallest possible value of x among the following?**

(a) 4

(b) 4.5

(c) 2.8

(d) 3.5

**Q10. Which of the following is correct?**

(a) Small “e” and capital “E” are both symmetrical about a line.

(b) Small “o” and capital “O” are both symmetrical about a line.

(c) Small “a” and capital “A” are both symmetrical about a line.

(d) Small “b” and capital “B” are both symmetrical about a line.

Solutions

S1. Ans.(b)

Sol.

Here, ∆ABC ≅ ∆DFE

Therefore, AC = DE, AB = DF and BC = EF

Also, ∠A = ∠D, ∠B = ∠F and ∠C = ∠E

Hence, AB = EF is not correct.

S2. Ans.(d)

Sol.

Two right-angled triangles are congruent, by the RHS congruence, if a side and the hypotenuse of one right-angled triangle are equal to the corresponding hypotenuse and side of another right-angled triangle.

S3. Ans.(a)

Sol. ∠BDC + ∠BDA = 180°

⇒ 119° + ∠BDA = 180°

⇒ ∠BDA = 180° – 119° = 61°

Now, ∠ABD = 180° – (61° + 62°)

= 180° – 123° = 57°

S4. Ans.(c)

Sol. For any polyhedron that does not intersect itself,

F + V – E = 2

Where

F is the number of faces, V is the number of vertices and E is the number of edges.

12 + 16 – E = 2

⇒ E = 12 + 16 – 2 = 26

S5. Ans.(c)

Sol. In ∆ABC,

80° + 60° + ∠ACB = 180°

(∵ Sum of three angles of a triangle is 180°)

∠ACB = 180° – 140° = 40°

∠DCE = ∠ACB = 40°

(∵ Vertically opposite angles)

Now in ∆DEC,

40° + 90° + y = 180°

y = 180° – 130° = 50°

S6. Ans.(a)

Sol.

We know that the longest side is always opposite the largest interior angle, and the shortest side is always opposite the least interior angle.

Therefore, ∠C > ∠A > ∠B

Hence, ∠A > ∠B

S7. Ans.(a)

Sol. PQ = TR, PR = TQ, and QR is the common side of two triangles. As all three sides of both triangles are equal, the given triangles are congruent.

The angle corresponding to ∠P is ∠T, and the angle corresponding to ∠Q is ∠R.

Therefore, ∆PQR and ∆TRQ are congruent.

S8. Ans.(d)

Sol. Two angles are supplementary in they add up to 180 degrees.

54° + 126° = 180°

Also, 172° + 8° = 180°

S9. Ans.(a)

Sol. We know that the largest side of a triangle is less than the sum of its two other sides.

Or

The least side of a triangle is greater than the difference of its other two sides.

∴ x > 10 – 6.5

⇒ x > 3.5 cm

In the given option, the smallest possible value of x greater than 3.5 is 4.

S10. Ans.(b)

Sol. Both small “o” and capital “0” are symmetrical about both x-axis and y-axis.